By abc we denote a three digit number with digits a, b, and c. Is abc

ABC is divisible by 3 => A+B+C needs to be divisible by 3

(1) The product of (a) times (b) is a number divisible by 3 => At least a or b is divisible by 3 => INSUFFICIENT
(2) c = 3 => INSUFFICIENT

(1) + (2) => At least 2 of a,b,c are divisible by 3, BUT NOT SURE a+b+c is divisible by 3

=> Answer E

MAGOOSH OFFICIAL SOLUTION:

Remember, the divisibility rule: if the sum of all the digits in a large number is divisible by 3, then the number itself is divisible by three.

The problem with statement #1, for any given values of a & b that work, the one’s digit, c, could be anything. For example, a = 1, and b = 3 satisfy this condition, so that would be all the numbers in the 130’s: {130, 131, 13, 133, 134, 135, 136, 137, 138, 139}. Clearly, some of these are divisible by 3, namely {132, 135, 138} and the others aren’t. This statement allows for multiple answers, so this statement, alone and by itself, is not sufficient.

The problem with statement #2 is that we now know the one’s digit, but the other two digits could be anything. Think about the first few possible numbers of this form: {103, 113, 123, 133, 143, 153, 163, 173, 183, 193, 203}. Of those, only (123, 153, 183) are divisible by three, and the others aren’t. In fact, you would be expected to know this, but as it happens, {103, 163, 173, 193} are prime numbers, not divisible by anything! This statement also allows for multiple answers, so this statement, alone and by itself, is not sufficient.

Combined statements: now, we have to obey both constraints. One number that obeys both constraints is 133, which is not divisible by 3. Another number that obeys both constraints is 333, which is divisible by 3. Even with both statements in place, we can find numbers that produce either a “yes” or a “no” to the prompt. Even together, the statements are not sufficient.

Answer = (E)

Excellent question..
Here 633 and 133 satisfy both the statements
hence E

The problem with statement #1, for any given values of a & b that work, the one's digit, c, could be anything. For example, a = 1, and b = 3 satisfy this condition, so that would be all the numbers in the 130's: {130, 131, 13, 133, 134, 135, 136, 137, 138, 139}. Clearly, some of these are divisible by 3, namely {132, 135, 138} and the others aren’t. This statement allows for multiple answers, so this statement, alone and by itself, is not sufficient.

The problem with statement #2 is that we now know the one's digit, but the other two digits could be anything. Think about the first few possible numbers of this form: {103, 113, 123, 133, 143, 153, 163, 173, 183, 193, 203}. Of those, only (123, 153, 183) are divisible by three, and the others aren’t. In fact, you would NOT be expected to know this, but as it happens, {103, 163, 173, 193} are prime numbers, not divisible by anything! This statement also allows for multiple answers, so this statement, alone and by itself, is not sufficient.

Combined statements: now, we have to obey both constraints. One number that obeys both constraints is 133, which is not divisible by 3. Another number that obeys both constraints is 333, which is divisible by 3. Even with both statements in place, we can find numbers that produce either a "yes" or a "no" to the prompt. Even together, the statements are not sufficient.

Answer = (E)

For divisibility of 3, a+b+c should be a multiple of 3.

Statement 1 is not sufficient as the number can be 331 or 333.

Statement 2 is also not sufficient as the number can be 223 or 333

Statement 1 +2 also not sufficient as number can be 133 or 333

So answer is E

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